Gauge Transformation of Lagrangians

Question

in my lecture of theoretic mechanics my prof introduced the invariance of the lagrangian under gauge transformation. I know how to do it and why i am allowed do it but i lack the knowledge of: why would i like to do it?

So my question is: What is the purpose of gauge transformation of a Lagrangian?

EDIT: I am sorry for the lack of definition of gauge transformation. I wanted to ask about transformations of the form: $$L'(q,\frac{d}{dt} q,t) = L(q,\frac{d}{dt} q,t) + \frac{d}{dt} f(q,t)$$ where $L$ is a Lagrangian and $L'$ is the gauge transformed Lagrangian. $f$ is a scalar-valued function. It could be that this is the result of a translation error. I study in Germany and here it is called "Eichtransformation". I did some search in the net and thought this would be translated to gauge transformation.

Answer 1

In theoretical mechanics, I presume you have transformations like $x \to x + \epsilon$ or $\phi \to \phi + \epsilon$. If you have a Lagrangian that does not depend on position (or angle) and only (angular) velocity $\dot x$ ($\dot \phi$), you will be able to derive conservation of (angular) momentum from this. Noether's theorem gives you conserved quantities based on a continous symmetry transformation.

In quantum field theory, gauge symmetries are used to introduce forces. Simple massive fermions can be described by $$ L = \int \mathrm d^3 x \, \mathrm dt \, \bar\psi(x, t) [\mathrm i \gamma^\mu \partial_\mu - m] \psi(x, t) \,. $$ There is a symmetry in this Lagrangian: You can rotate $\psi \to \mathrm e^{\mathrm i \alpha} \psi$. This will also rotate the complex conjugate $\bar\psi \to \bar\psi e^{-\mathrm i \alpha}$. Since that is just a complex number, you can just commute that with the square bracket, the two terms multiply to 1, and $L$ is left invariant. That is a global symmetry that will lead to charge conservation.

The quantum mechanical interpretation is not changed by this, because the observable is always something like $|\psi|^2$ where the complex chase $\mathrm e^{\mathrm i \alpha}$ would square away. We can therefore change the complex phase of the wavefunction without changing the physics that it describes. Currently we just change the phase for the whole space. But if that phase does not matter, couldn't we also just change it to a different value of $\alpha$ for every point in spacetime? Let us try. We then have $$\psi \to \mathrm e^{\mathrm i \alpha(x)} \psi \,, \qquad \bar\psi \to \bar\psi e^{-\mathrm i \alpha(x)} \,.$$ In the expression $\bar\psi\psi$, the two complex phases will just cancel each other. But in the original expression, we have a partial derivative. With the chain rule, we will obtain a factor $\mathrm i \partial_\mu \alpha$. This breaks the local gauge invariance.

The trick is to add a term canceling that. What happens mathematically is that you have a non-trivial fiber bundle and therefore need to introduce a connection instead of a normal partial derivative. One introduces a vector field $A_\mu$ to the derivative. That field must have the property of transforming like $$A_\mu \to A_\mu + \mathrm i \frac{\partial \alpha}{\partial x^\mu} \,.$$

The resulting Lagrangian then is this: $$ L = \int \mathrm d^3 x \, \mathrm dt \, \bar\psi(x, t) \left[\mathrm i \gamma^\mu \left(\partial_\mu - A_\mu \right) - m\right] \psi(x, t) \,. $$ When we perform the gauge transformation, the unwanted term is cancelled cleanly: $$ L = \int \mathrm d^3 x \, \mathrm dt \, \bar\psi(x, t) \left[\mathrm i \gamma^\mu \left(\partial_\mu - \mathrm i \frac{\partial \alpha}{\partial x^\mu} + A_\mu - \mathrm i \frac{\partial \alpha}{\partial x^\mu}\right) + - m\right] \psi(x, t) \,. $$

One calls this additional vector the connection coefficient (math) or connection (physics). The whole parentheses is called connection (math) or gauge covariant derivative (physics). This vector field $A_\mu$ will then describe photons in your theory.

So gauge invariance is a very powerful tool to introduce forces into a quantum theory of particles. For classical mechanics, you can get conserved quantities.

Answer 2

You should have specified, which gauge transformations where introduced in your course. In physics, quite a large number of examples go with this name. So, I am going to try to guess.

In classical mechanics, were you study non-relativistic motion of a particle, in most cases there only symmetries, that are present for the equations of motion are "global", that is universal over space. For example, the free particle action is invariant under shifts in space $x(t)\to x(t)+x_0$ (for each particle), or shifts in time $x(t)\to x(t+t_0)$. Both are very useful - first leads to momentum conservation, second to energy conservation.

However, one situation in classical mechanics is rather peculiar - that is the case of motion of a particle in electromagnetic field. While it is rather easy to write the Newtons equation of motion for such a particle in terms of well known electric and magnetic fields $E$ and $B$ $$ m\ddot{\mathbf{x}} = q\mathbf{E}+q\dot{\mathbf{x}}\times\mathbf{B}, $$ it turns out that the task to write an action, leading to these equations, is rather hard. To do so, it is required to introduce the scalar and vector potentials, $\phi$ and $\mathbf{A}$, with the action $$ S=\int dt \left({1\over2} \,m \,\dot{\mathbf{x}}^2-e\, \mathbf{A}(t, \mathbf{x})\cdot \dot{\mathbf{x}}-e \,A_0(t, \mathbf{x})\right). $$ This would lead to the proper equations of motion, given $\mathbf{E}=-\nabla A_0$, $\mathbf{B}=-\nabla\times\mathbf{A}$ (for static fields). However, it turns that this is not a unique choice! Any functions $A_0'(x,t)=A_0+d\alpha(x,t)/dt$, $\mathbf{A}'(x,t)=\mathbf{A}+\nabla\alpha(x,t)$ will lead to exactly the same equations of motion for the particle. Here $\alpha(x,t)$ is an arbitrary function of time and space - "gauge" transformation.

For now, this is a mathematical peculiarity. One could just use only Newton's laws of motion, electric and magnetic field, and never bother about vector potential and least acction principle. At the end, only gauge invariant quantities, $\mathbf{E}$ and $\mathbf{B}$ are really measured and physical! However, it turns out, that the formulation in terms of $\mathbf{A}$ and $A^0$ is very deep and useful, though it has this funny "gauge invariance" feature. Really, what happens that in fact the statement, that only gauge invariant quantities are observed goes beyond $E$ and $B$. Unfortunately, you can not really see that in classical mechanics, but in quantum theory this leads to real physical effects. This is the Aharonov–Bohm effect. If in a space both $E=B=0$, it is still possible to have nonzero $\mathbf{A}$, if $\nabla\times\mathbf{A}=0$. However, one can make configurations that an integral over a noncontractable loop $$ \oint \mathbf{A} d\mathbf{x} \neq 0 $$ is not zero, though the magnetic field is zero everywhere along the loop. This integral is actually equal to the magnetic flux through the loop (say, there is an infinitely long solenoid inside the loop). And in quantum mechanics this leads to an interference pattern between particles that never enter the region with magnetic field, but only travel in the space around the solenoid! Thus, the description in terms of the scalar and vector potential is more complete, than with the magnetic and electric fields, at the price of an extra complication of gauge invariance.

Later on, in the quantum field theory, it turns out that the gauge principle governs all the fundamental forces of nature, not only electromagnetism.

As a side note, there is one more gauge transformation that may appear in mechanics, that is the reparametrisation invariance in teh world line action for a relativistic particle (a one dimensional analogue of a better known Polyakov action for a string), but this is probably better to leave until the study of general relativity and string theory.